Three capacitors of capacities `8muF,8muFand4muF` are connected in a series and a potential difference of 120 volt is maintained across the combination. Calculate the charge on capacitor of capacity `4muF`.

Explanation

Let the capacitors be `C_(1)=8muF,C_(2)=8muFandC_(3)=4muF`.
All the capacitors are connected in series, so, the equivalent capacitance `C_(s)` can be calculated as
`(1)/(C_(s))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))`
`(1)/(C_(s))=(1)/(8)+(1)/(8)+(1)/(4)`
`(1)/(C_(s))=(1+1+2)/(8)`
`C_(s)=2muF`
In series combination,
`Q_(1)=Q_(2)=Q_(3)`
But `Q=C_(s)V`
`:.` Charge on capacitor of capacity `4muF` can be calculated as
`Q=2xx10^(-6)xx120`
`=240xx10^(-6)C=240muC`